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Question number: 4

Question

A singly ionized helium atom in an excited state (n = 4) emits a photon of energy 2.6 eV. Given that the ground state energy of hydrogen atom is -13.6 eV, the energy Equation and quantum number (n) of the resulting state are respectively,

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Question number: 7

Question

If the wavelength of photon emitted due to transition of electron from third orbit to first orbit in a hydrogen atom is λ, then the wavelength of photon emitted due to transition of electron from fourth orbit to second orbit will be –